Solving a system of equations involving an absolute value Solve the following system:
$$
\begin{cases}
\text{a}\cdot\text{c}+\text{b}\cdot\text{d}\cdot\epsilon^2=\epsilon\cdot\left(\text{b}\cdot\text{c}-\text{a}\cdot\text{d}\right)\\
\\
\left|\epsilon\right|=\left|-\frac{\text{c}}{\text{d}}\right|
\end{cases}
$$
I don't know how to proceed?!

With the help of the comments and the answer I got:
$$
\begin{cases}
\text{a}\cdot\text{c}+\text{b}\cdot\text{d}\cdot\epsilon^2=\epsilon\cdot\left(\text{b}\cdot\text{c}-\text{a}\cdot\text{d}\right)\\
\\
\epsilon=\pm\frac{\text{c}}{\text{d}}
\end{cases}\to\text{a}\cdot\text{c}+\text{b}\cdot\text{d}\cdot\frac{\text{c}^2}{\text{d}^2}=\pm\frac{\text{c}}{\text{d}}\cdot\left(\text{b}\cdot\text{c}-\text{a}\cdot\text{d}\right)
$$
 A: A case disctinction helps. Assume first that $\epsilon=c/d$. Then we obtain $ac+bc^2/d=c^2b/d-ca$, hence $2ac=0$. Both cases $a=0$ or $c=0$ are easily solved.
A: If $c/d$ is positive, you will have $|-c/d|=c/d$. Thus
$$
ac+bd\frac{c^2}{d^2}=\dfrac{c}{d}(bc-ad),
$$
or 
$$
ac+\frac{bc^2}{d}=\dfrac{bc^2}{d}-ac,
$$
or 
$$ac=0.$$
Now if $c/d$ is negative, you will have $|-c/d|=-c/d$. Thus
$$
ac+bd\frac{c^2}{d^2}=-\dfrac{c}{d}(bc-ad),
$$
or 
$$
ac+\frac{bc^2}{d}=-\dfrac{bc^2}{d}+ac,
$$
or 
$$bc^2/d=0.$$
A: We are given
$$
\left\{ \begin{gathered}
  bd\varepsilon ^{\,2}  - \left( {bc - ad} \right)\varepsilon  + ac = 0 \hfill \\
  \varepsilon  =  \pm \frac{c}
{d} \hfill \\ 
\end{gathered}  \right.
$$
So we have the equation of a vertical parabola in $\varepsilon$, whose intercepts with the $x$ axis
must be symmetrical (vs. $\varepsilon=0$).
Thus the parabola must be symmetrical as well, i.e. the coefficient
of the $\varepsilon$ term must be null, which gives:
$$
\begin{gathered}
  \left\{ \begin{gathered}
  bd\varepsilon ^{\,2}  - \left( {bc - ad} \right)\varepsilon  + ac = 0 \hfill \\
  \varepsilon  =  \pm \frac{c}
{d} \hfill \\ 
\end{gathered}  \right.\quad  \Rightarrow \quad \left\{ \begin{gathered}
  bc - ad = 0 \hfill \\
  b\frac{{c^{\,2} }}
{d} + ac = 0 \hfill \\ 
\end{gathered}  \right.\quad  \Rightarrow  \hfill \\
   \Rightarrow \quad \left\{ \begin{gathered}
  bc - ad = 0 \hfill \\
  d \ne 0 \hfill \\
  \left( {bc + ad} \right)c = 0 \hfill \\ 
\end{gathered}  \right. \Rightarrow \quad \left\{ \begin{gathered}
  c = 0 \hfill \\
  d \ne 0 \hfill \\
  a = 0 \hfill \\
  \forall b \hfill \\ 
\end{gathered}  \right. \hfill \\ 
\end{gathered} 
$$
